A 16Ω wire is bend to form a square loop. A 9V supply having internal resistance of 1Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is __________ x 10?¹ V.
A 16Ω wire is bend to form a square loop. A 9V supply having internal resistance of 1Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is __________ x 10?¹ V.
Each side of the square has a resistance of 16/4 = 4Ω.
The side AC has the 9V, 1Ω source.
The other three sides (AB, BD, DC) form a path with resistance 4+4+4 = 12Ω.
This is in parallel with the side AC (4Ω).
Total resistance of the loop part: (12*4)/ (12+4) = 48/16 = 3Ω.
Total resistance of the circuit
Similar Questions for you
Ohm's law is valid if I depends on V' linearly.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...

Physics Current Electricity 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering




